Now upon individually converting these fractions back into power series, in short, what I obtained is a power series with coefficients $\frac{(n+3)^2}{12} + x_n$ for $\frac{1}{3} \leq x_n \leq \frac{2}{3}$.

This was expected, since my book states that $r(n)$ is given by the integer closest to $\frac{(n+3)^2}{12}$. What I don't understand is how can we conclude such a close correspondence between $r(n)$ and $\frac{(n+3)^2}{12} + x_n$? Just because two power series converge to the same closed form function, can't their coefficients not still be very different for any particular $n$? I know Taylor series are unique, but I don't think my latter power series is a Taylor series, and even if the power series expansion of a function is always unique, it still has that extra $x_n$ added to or subtracted from each coefficient. What is going on here?

Edit: Ok so $r(n)$ is supposed to be the number of ordered triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$. I formed this by employing the Cauchy product $(\sum z^n)(\sum z^{2n})(\sum z^{3n})$.

2 Answers
2

You have two power series expansions of the same function, one with coefficients $r(n)$, one with coefficients $(1/12)(n+3)^2+x_n$. By the uniqueness of power series expansions, $r(n)=(1/12)(n+3)^2+x_n$. Unless, of course, you have edited your post several more times while I was typing this.